Thread: Good with integrals?

A 0.1-F capacitor measures 150V across it. At t=0, the capacitor is connected to a source that sends current i=16t/square root of (4t^2+9) amperes through the circuit. Find the voltage across the capacitor when t=2s.

I think you use this equation, but im not sure.
Vc=(1/C)xintegral of ixdt

You're completely right.

Basically, you know the Capacitance, and you're giving the function of current with respect to time. Since $\displaystyle I=\frac{\delta Q}{\delta t}$ you can just Integrate your I function with respect to t.

$\displaystyle \int_0^2{\frac{16t}{\sqrt{4t^2+9}}}$

Just use a simple u substitution so that

$\displaystyle u=4t^2+9$ and $\displaystyle \delta u=8t$

From there, you can just integrate the function as $\displaystyle 2\int_0^2{\frac{1}{\sqrt{u}}}$

and you will end up with the answer $\displaystyle 4\sqrt{u}$

Change your $\displaystyle u$ back into $\displaystyle 4t^2+9$

Once you plug in your limits of 0 and 2 you will get $\displaystyle \frac{4\sqrt{25}}{C}$

You know that $\displaystyle C=.1$ so your final result should be $\displaystyle V=200V$